# arithmetic-geometric series

• Feb 5th 2007, 05:01 PM
kcsteven
arithmetic-geometric series
The general arithmetic-geometric series of n terms is a series of the form:
a + (a+d)r + (a + 2d)r^2 + ... +[a + (n - 1)d]r^n-1 Where the first factors of each term form an arithmetic series and the second factors form a geometric series. What technique could I use to derive a formula for the sum of this series? Could I let S denote the sum of the general geometric series of n terms. Multiply S by r and write the series corresponding to the difference between S and rS. How would I use this expression to deducde the formula for the sum of the general geometric series? :eek:
• Feb 5th 2007, 07:02 PM
AlvinCY
I think you'll find this useful:

PlanetMath: arithmetic-geometric series
• Feb 5th 2007, 09:52 PM
Soroban
Hello, Keith!

Quote:

The general arithmetic-geometric series of $\displaystyle n$ terms is a series of the form:
. . $\displaystyle S \:=\:a + (a+d)r + (a + 2d)r^2 + \cdots +[a + (n - 1)d]r^{n-1}$
where the first factors of each term form an arithmetic series
and the second factors form a geometric series.

What technique could I use to derive a formula for the sum of this series?
Could I let $\displaystyle S$ denote the sum of the general geometric series of $\displaystyle n$ terms?
Multiply $\displaystyle S$ by $\displaystyle r$ and then write the series corresponding
. . to the difference between $\displaystyle S$ and $\displaystyle rS$ ? . Yes ... a great game plan!

How would I use this to deduce the formula for the sum of the general series?

The original series:
. . . $\displaystyle S\:=\:a + (a+d)r + (a+2d)r^2 + (a+3a)r^3 + \cdots + [a+(n-1)d]r^{n-1}$

Multiply by $\displaystyle r\!:$
. . .$\displaystyle rS\:=\qquad\quad ar \quad+ \quad(a+d)r^2 + (a+2d)r^3 + \cdots$ $\displaystyle + [a+(n-2)d]r^{n-1} + [a + [n-1]d]r^n$

Subtract: .$\displaystyle S - rS\:=\;a + \underbrace{dr + dr^2 + dr^3 + \cdots + dr^{n-1}}_{\text{geometric series}} - [a + (n-1)d]r^n$

. . The geometric series has first term $\displaystyle dr$, common ratio $\displaystyle r$, and $\displaystyle n-1$ terms.
. . . . Its sum is: .$\displaystyle dr\,\frac{1-r^{n-1}}{1-r}$

So we have: .$\displaystyle (1 - r)S\;=\;a +dr\,\frac{1 - r^{n-1}}{1-r} - [a + (n-1)d]r^n$

. . $\displaystyle (1-r)S\:=\:\frac{a(1-r) + dr(1 - r^{n-1}) - (1-r)[a + (n-1)d]r^n}{1 - r}$

Therefore: .$\displaystyle S \:=\:\frac{a(1-r) + dr(1 - r^{n-1}) - (1-r)[a + (n-1)d]r^n}{(1-r)^2}$